A Study of Some Integral Problems Using Maple

A Study of Some Integral Problems Using Maple

Chii-Huei Yu

Department of Management and Information, Nan Jeon University of Science and Technology, Tainan City, 73746, Taiwan *Corresponding Author: [email protected]

Copyright © 2014 Horizon Research Publishing All rights reserved.

Abstract

This paper takes the mathematical software Maple as the auxiliary tool to study four types of integrals. We can obtain the Fourier series expansions of these four types of integrals by using integration term by term theorem. On the other hand, we provide two examples to do calculation practically. The research methods adopted in this study involved finding solutions through manual calculations and verifying these solutions by using Maple.

Keywords

Integrals, Fourier Series Expansions, Integration Term By Term Theorem, Maple

1. Introduction

In calculus and engineering mathematics courses, we learnt many methods to solve the integral problems including change of variables method, integration by parts method, partial fractions method, trigonometric substitution method, and so on. In this paper, we mainly study the following four types of integrals which are not easy to obtain their answers using the methods mentioned above.

∫

sin[rcos(at+b)]cosh[rsin(at+b)]dt (1)

∫

cos[rcos(at+b)]sinh[rsin(at+b)]dt (2)

∫

cos[rcos(at+b)cosh[rsin(at+b)]dt (3)

∫

sin[rcos(at+b)sinh[rsin(at+b)]dt (4) Where r,a,bare real numbers,a≠0. We can obtain the Fourier series expansions of these four types of integrals by using integration term by term theorem; these are the major results of this paper (i.e., Theorems 1, 2). As for the study of related integral problems can refer to [1-15]. On the other hand, we propose some integrals to do calculation practically. The research methods adopted in this study involved finding solutions through manual calculations and verifying these solutions by using Maple. This type of research method not only allows the discovery of calculation errors, but also helps modify the original directions of thinking from manual and Maple calculations.

For this reason, Maple provides insights and guidance regarding problem-solving methods.

2. Main Results

Firstly, we introduce a notation and some formulas used in this study.

2.1. Notation

Let be a complex number, where i= −1, are real numbers. We denote the real part of by , and the imaginary part of by

.

2.2. Formulas

2.2.1. Euler's formula

θ θ

θ _{cos isin}

ei = + , where

θ

is any real number. 2.2.2. DeMoivre's formula

θ

θ

θ

θ

isin )n cosn isinn

(cos + = + , where n is

any integer,

θ

is any real number. 2.2.3. ([16, p25])

v u i v u iv

u ) sin cosh cos sinh

sin( + = + _{, where}

v

u, are real numbers. 2.2.4. ([16, p25])

v u i v u iv

u ) cos cosh sin sinh

cos( + = − , where u,v

are real numbers. 2.2.5. ([16, p63])

1 2

0 (2 1)!

1 ) 1 (

sin ∞ +

= +

−

=

∑

k

k

k _z

k

z , where z is any

complex number. 2.2.6. ([16, p63])

k k

k _z

k

z 2

0 (2 )!

1 ) 1 ( cos

∑

∞

=

−

= , where z is any complex

number.

ib a z= +

b

a, a z

)

Re(z b z

Next, we introduce an important theorem used in this paper.

2.3. Integration term by term theorem ([17, p269])

Suppose

{

g_k(t)

}

∞_k=0 is a sequence of Lebesgue integrable functions defined on an inteval [c,d] . If

dt t g

k d c k

∑ ∫

∞

=0 ( )

is convergent, then _cd g t dt

k k

∫ ∑

∞

=0 ( )

∑ ∫

∞

=

=

0 ( )

k d

c gk t dt.

The following is the first major result of this study, we obtain the Fourier series expansions of the integrals (1) and (2).

2.4. Theorem 1

Assume r,a,b,c are real numbers, a≠0. Then there exists a constant C₁such that for all x∈R, the integral

∫

_cxsin[rcos(at+b)]cosh[rsin(at+b)]dt

1 1

2

0( 1) (2 1)!(2 1)sin[(2 1)( )]

1 _{k ax b C}

k k

r a

k

k

k _{+ + +}

+ +

− ⋅

= ∞ +

=

∑

(5) And there exists a constant C₂such that for all x∈R, the integral

∫

_cxcos[rcos(at+b)]sinh[rsin(at+b)]dt

2 1

2

0( 1) (2 1)!(2 1)cos[(2 1)( )]

1 _{k ax b C}

k k

r a

k

k

k _{+ + +}

+ +

− ⋅ −

= ∞ +

=

∑

(6) 2.4.1. Proof

Because

)] sin(

cosh[ )] cos(

sin[r ax+b r ax+b

)]} (

exp

Re{sin[r i ax+b

=

(By Formula 2.2.3)

    

  

+ +

−

= ∞ +

=

∑

2 1

0 (2 1)![ exp ( )]

1 ) 1 (

Re k

k

k _{r i ax b}

k (Using Formula 2.2.5)

    

  

+ +

+ −

= ∞ +

=

∑

exp [(2 1)( )]

)! 1 2 ( ) 1 (

Re 2 1

0 k i k ax b

r k

k

k

(By DeMoivre's formula)

] ) )( 1 2 cos[( )! 1 2 ( ) 1 (

0

1 2

∑

∞

=

+

+ +

+ −

= k

k

k _{k ax b}

k r

(7)

(By Euler's formula)

Thus, for all x∈R, the integral

∫

_cxsin[rcos(at+b)]cosh[rsin(at+b)]dt

∫ ∑

∞

=

+

+ +

+ −

= _cx k

k

k _{k at b dt}

k r 0

1 2

)] )(

1 2 cos[( )! 1 2 ( ) 1 (

∫

∑

+ +

+ −

= ∞ +

=

x c k

k

k _{k at b dt}

k

r _{cos[(2 1)( )]}

)! 1 2 ( ) 1

( 2 1

0

(By integration term by term theorem)

1 1

2

0( 1) (2 1)!(2 1)sin[(2 1)( )]

1 _{k ax b C}

k k

r a

k

k

k _{+ + +}

+ +

− ⋅

= ∞ +

=

∑

Where C₁is some constant.

On the other hand, using Euler's formula, DeMoivre's formula and Formula 2.2.3, 2.2.5, we have

)] sin(

sinh[ )] cos(

cos[r ax+b r ax+b

)]} (

exp

Im{sin[r i ax+b

=

] ) )( 1 2 sin[( )! 1 2 ( ) 1 (

0

1 2

∑

∞

=

+

+ +

+ −

= k

k

k _{k ax b}

k r

(8)

Therefore, by integration term by term theorem, we can show that there exists a constant C₂ such that for all

R

x∈ , the integral

∫

_cxcos[rcos(at+b)]sinh[rsin(at+b)]dt

2 1

2

0( 1) (2 1)!(2 1)cos[(2 1)( )]

1 _{k ax b C}

k k

r a

k

k

k _{+ + +}

+ +

− ⋅ −

= ∞ +

=

∑

q.e.d.

2.5. Remark 1

In Theorem 1, because for each x∈R,

dt b at k

k r

k x c

k k

∑ ∫

∞

=

+

+ +

+ −

0

1 2

)] )(

1 2 cos[( )! 1 2 ( ) 1 (

∑

∞

∫

=

+

+ + +

+ ⋅

≤

0

1 2

)] )(

1 2 cos[( )

1 2 ( )! 1 2 ( 1

k

x c k

dt b at k k

k r a

2 1

0

1

(2 1)!(2 1)

k

k

r x c

a k k

+ ∞

=

−

≤ ⋅ < ∞

+ +

∑

It follows that we can use integration term by term theorem to show that (5) holds. The same reason that we can prove (6) by using integration term by term theorem.

Next, we determine the Fourier series expansions of the integrals (3) and (4).

2.6. Theorem 2

exists a constant C₃such that for all x∈R, the integral

∫

_cxcos[rcos(at+b)]cosh[rsin(at+b)]dt

3 2

1( 1) (2 )!2 sin[(2 )( )]

1 _{k ax b C}

k k r a

x k

k

k _{+ +}

− ⋅ +

=

_∑

∞

=

(9) And there exists a constant C₄such that for all x∈R, the integral

∫

_cxsin[rcos(at+b)]sinh[rsin(at+b)]dt

4 2

1( 1) (2 )!2 cos[(2 )( )]

1 _{k ax b C}

k k r a

k

k

k _{+ +}

− ⋅ =

_∑

∞

= (10)

2.6.1. Proof

By Euler's formula, DeMoivre's formula and Formula 2.2.4, 2.2.6, we have

)] sin(

cosh[ )] cos(

cos[r ax+b r ax+b

)]} (

exp

Re{cos[r i ax+b

=

] ) )( 2 cos[( )! 2 ( ) 1 (

0

2

∑

∞

=

+ −

= k

k

k _{k ax b}

k r

] ) )( 2 cos[( )! 2 ( ) 1 ( 1

1

2

∑

∞

=

+ −

+ =

k

k

k _{k ax b}

k

r ₍₁₁₎

Thus, for all x∈R, the integral

∫

_cxcos[rcos(at+b)]cosh[rsin(at+b)]dt

∫

∑

− +

+ −

= ∞

=

x c k

k

k _{k at b dt}

k r c

x cos[(2 )( )] )!

2 ( ) 1 ( 2

1

(By integration term by term theorem)

3 2

1( 1) (2 )!2 sin[(2 )( )]

1 _{k ax b C}

k k r a

x k

k

k _{+ +}

− ⋅ +

=

_∑

∞

=

Where C₃is some constant.

Similarly, by Euler's formula, DeMoivre's formula and Formula 2.2.4, 2.2.6, we obtain

)] sin(

sinh[ )] cos(

sin[r ax+b r ax+b

)]} (

exp

Im{cos[r i ax+b

−

=

] ) )(

2 sin[( )! 2 ( ) 1 (

1

2

∑

∞

=

+ −

− =

k

k

k _{k ax b}

k r

(12)

By integration term by term theorem, it follows that for all R

x∈ , the integral

∫

_cxsin[rcos(at+b)]sinh[rsin(at+b)]dt

4 2

1( 1) (2 )!2 cos[(2 )( )]

1 _{k ax b C}

k k r a

k

k

k _{+ +}

− ⋅ =

_∑

∞

=

Where C₄is some constant. q.e.d.

2.7. Remark 2

In Theorem 2, the reason that we can use integration term by term theorem to prove (9) and (10) is the same as Remark 1.

3. Examples

In the following, for the four types of integrals in this study, we propose some integrals and use Theorems 1, 2 to determine their Fourier series expansions. In addition, we evaluate some definite integrals and employ Maple to calculate the approximations of these definite integrals and their solutions for verifying our answers.

3.1. Example 1

In Theorem 1, taking r=5,a=2,b=

π

/3into (5), we obtain the following integral

dt t

t

x c

∫



  

 

   

  ₊ 

  

 

   

  ₊

3 2 sin 5 cosh 3

2 cos 5

sin

π

π

1 1

2

0 (2 1)!(2 1)sin (2 1) 2 3

5 )

1 ( 2

1 _{k x C}

k k

k

k

k ₊

   

 

   

  ₊ + +

+ −

⋅

= ∞ +

=

∑

π

(13) Thus, we can determine the definite integral from t=π/12

to t=

π

/6,

∫

ππ//126sin_5cos_2t+

π

₃__cosh_5sin_2t+

π

₃__dt

3 ) 2 4 ( sin ) 1 2 ( )! 1 2 (

5 )

1 ( 2

1 2 1

0

π

+ +

+ −

⋅

= ∞ +

=

∑

_k k _k k k

k

2 ) 1 2 ( sin ) 1 2 ( )! 1 2 (

5 )

1 ( 2

1 2 1

0

π

+ +

+ −

⋅

− ∞ +

=

∑

_k k _k k k

k ₍₁₄₎

We use Maple to verify the correctness of (14).

>evalf(int(sin(5*cos(2*t+Pi/3))*cosh(5*sin(2*t+Pi/3)),t= Pi/12..Pi/6),14);

>evalf(1/2*sum((-1)^k*5^(2*k+1)/((2*k+1)!*(2*k+1))*s in((4*k+2)*Pi/3),k=0..infinity)-

1/2*sum((-1)^k*5^(2*k+1)/((

2*k+1)!*(2*k+1))*sin((2*k+1)*Pi/2),k=0..infinity),14);

(= −1), it is because Maple calculates by using special functions built in. Both the imaginary parts of the above answers are zero, so can be ignored.

On the other hand, in Theorem 1, taking

4 / 3 , 5 ,

2 = =−

π

= a b

r into (6), we have the

following integral

dt t

t

x

c _

          −             −

∫

sinh 2sin 5 3₄

4 3 5 cos 2

cos

π

π

2 1 2 0 4 3 5 ) 1 2 ( cos ) 1 2 ( )! 1 2 ( ) 2 ( ) 1 ( 5

1 _{k x C}

k k k k k ₊             ₋ + + + − ⋅ − = ∞ + =

∑

π (15) Hence, we can determine the definite integral from

10 /

π

=

t to t =2

π

/5,

∫

            −             − 5 / 2 10

/ cos 2cos 5 3₄ sinh 2sin 5 3₄

π

π t

π

t

π

dt

4 ) 3 6 ( cos ) 1 2 ( )! 1 2 ( ) 2 ( ) 1 ( 5

1 2 1

0

π

+ + + − ⋅ − = ∞ + =

∑

_k k_k k k k 4 ) 1 2 ( cos ) 1 2 ( )! 1 2 ( ) 2 ( ) 1 ( 5

1 2 1

0

π

+ + + − ⋅ + ∞ + =

∑

_k k_k k k

k

(16) Next, we use Maple to verify the correctness of (16). >evalf(int(cos(sqrt(2)*cos(5*t-3*Pi/4))*sinh(sqrt(2)*sin( 5*t-3*Pi/4)),t=Pi/10..2*Pi/5),14); >evalf(-1/5*sum((-1)^k*sqrt(2)^(2*k+1)/((2*k+1)!*(2*k +1 ))*cos((6*k+3)*Pi/4),k=0..infinity)+1/5*sum((-1)^k*sqrt (2)^(2*k+1)/((2*k+1)!*(2*k+1))*cos((2*k+1)*Pi/4),k=0.. infinity),14);

Also, both the imaginary parts of the above answers obtained by Maple are zero, so can be ignored.

3.2. Example 2

In Theorem 2, taking r=4/3,a=7,b=5

π

/6into (9) , we can determine the following integral

dt t

t

x

c _

          ₊             ₊

∫

sin 7 5₆

3 4 cosh 6 5 7 cos 3 4

cos π π

3 2 1 6 5 7 ) 2 ( sin 2 )! 2 (4/3) ( ) 1 ( 7

1 _{k x C}

k k x k k k ₊             ₊ − ⋅ + =

_∑

∞ = π (17) Thus, we obtain the definite integral from t=

π

/14 to

7 / 2

π

= t ,

∫

            +             + 7 / 2 14

/ cos ₃4cos 7 5₆ cosh 4₃sin 7 5₆

π

π t

π

t

π

dt

3 5 sin 2 )! 2 ( ) 3 / 4 ( ) 1 ( 7 1 14 3 2 1

π

π

k k k k k k

∑

∞ = − ⋅ + = 3 2 sin 2 )! 2 ( ) 3 / 4 ( ) 1 ( 7 1 2 1

π

k k k k k k

∑

∞ = − ⋅

− (18)

We use Maple to verify the correctness of (18) as follows: >evalf(int(cos(4/3*cos(7*t+5*Pi/6))*cosh(4/3*sin(7*t+5 *Pi/6)),t=Pi/14..2*Pi/7),14); >evalf(3*Pi/14+1/7*sum((-1)^k*(4/3)^(2*k)/((2*k)!*(2* k))*sin(5*k*Pi/3),k=1..infinity)-1/7*sum((-1)^k*(4/3)^(2*k )/((2*k)!*(2*k))*sin(2*k*Pi/3),k=1..infinity),14);

Also, both the imaginary parts of the above answers obtained by Maple are zero, so can be ignored.

On the other hand, in Theorem 2, if taking

3 / 2 , 4 ,

9 = =−

π

= a b

r into (10), we obtain the

following integral

dt t

t

x

c _

          ₋             ₋

∫

sinh 9sin 4 2₃

3 2 4 cos 9

sin

π

π

4 2 1 3 2 4 2 cos 2 )! 2 ( 9 ) 1 ( 4

1 _{k x C}

k k k k k ₊             ₋ − ⋅ =

_∑

∞ =

π

(19) Therefore, we have the definite integral from t=

π

/6 to

3 /

π

= t ,

∫

            −             − 3 / 6

/ sin 9cos 4 2₃ sinh 9sin 4 2₃

π

π t

π

t

π

dt

3 4 cos 2 )! 2 ( 9 ) 1 ( 4 1 2 1

π

k k k k k k

∑

∞ = − ⋅ = k k k k k 2 )! 2 ( 9 ) 1 ( 4 1 2 1

∑

∞ = − ⋅

− (20)

Using Maple to verify the correctness of (20) as follows: >evalf(int(sin(9*cos(4*t-2*Pi/3))*sinh(9*sin(4*t-2*Pi/3)),t =Pi/6..Pi/3),14);

The imaginary parts of the above answers obtained by Maple are either zero or very small, so can be ignored.

4. Conclusion

From the above discussion, we know the integration term by term theorem plays a significant role in the theoretical inferences of this study. In fact, the application of this theorem is extensive, and can be used to easily solve many difficult problems; we endeavor to conduct further studies on related applications. On the other hand, Maple also plays a vital assistive role in problem-solving. In the future, we will extend the research topic to other calculus and engineering mathematics problems and solve these problems by using Maple. These results will be used as teaching materials for Maple on education and research to enhance the connotations of calculus and engineering mathematics.

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